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Lesson Video: Analyzing Combination Circuits Physics • 9th Grade

In this video, we will learn how to determine the currents through and voltages across parts of circuits that contain resistors both in series and parallel.

16:04

Video Transcript

In this video, weโ€™ll be learning about the characteristics of combination circuits, which are circuits that contain resistors in both series and parallel. We will begin by learning how to determine the equivalent resistance of a combination circuit. But first, we need to refresh our memory on how to find the equivalent resistance of resistors only in series or only in parallel.

Resistors in series are connected along a single conductive path, as shown in the diagram below. Resistors in parallel are connected along multiple conductive paths such that the current is split amongst them, as shown in the diagram below.

To determine the equivalent resistance of a series circuit, the value of one resistor that could replace the entire circuit, also known as the total resistance, we add each of the resistors together. For the diagram that we drew, the total resistance will be equal to ๐‘… one plus ๐‘… two plus ๐‘… three. If we had more resistors in series, we would need to take those values into account and add them in our equation. In a series circuit, the total resistance will always be larger than any of the individual resistors.

For a parallel circuit, finding the total resistance is a bit trickier. One over the total resistance is equal to one over ๐‘… one plus one over ๐‘… two plus one over ๐‘… three. The value of any additional resistor added in parallel would need to be taken into account. The total resistance for a parallel circuit will always be smaller than the smallest resistor. Letโ€™s apply the rules for finding equivalent resistance for series and parallel circuits to a combination circuit.

To determine the equivalent resistance of a combination circuit, we need to simplify our circuit down to either a simple series or a simple parallel circuit. Looking at our combination circuit, we can see that it would be a simple series circuit if not for these two resistors which are in parallel. This means that if we replace those two resistors with one resistor of equivalent resistance, we would have a simple series circuit.

Letโ€™s draw out the simplified circuit. In our simplified circuit, we still have ๐‘… one of five ohms and ๐‘… four of four ohms, just as we did in the circuit up above. But this time we have only one resistor where before we had two resistors in parallel. Now, we need to find the equivalent resistance of those two resistors in parallel using the equation for finding resistance of resistors in parallel.

Remember that one over the equivalent resistance is equal to one over the first resistance plus one over the second resistance plus one over the third resistance so on and so forth for however many resistors that are in parallel. Plugging in our values, the left side of the equation stays the same with one over the equivalent resistance. And the right side of the equation, since resistor two has a value of six ohms, we have one divided by six ohms plus, resistor three has a value of three ohms, one divided by three ohms.

When adding fractions, we need to find the least common denominator. In this case, it would be six ohms. We have to multiply one over three ohms by two over two so that it becomes two over six ohms. One over six ohms plus two over six ohms equals three over six ohms, which can simplify down to one over two ohms. To solve for the equivalent resistance, we need to multiply both sides of our equation by two ohms and equivalent resistance. On the left side of the equation, ๐‘… equivalent will cancel out. And on the right side of the equation, two ohms will cancel out, leaving us with an equivalent resistance of two ohms.

We can, therefore, say that our single resistor ๐‘… is equal to two ohms, which is the equivalent resistance of the three-ohm and the six-ohm resistors in parallel. We use the letter ๐‘… to represent our resistor so that we donโ€™t get confused when we go to find the equivalent resistance for the whole circuit. Now, we have a simple series circuit with three resistors: ๐‘… one, ๐‘…, and ๐‘… four. We can use the equation for resistors in series to determine the total resistance of the circuit.

Remember that the equivalent resistance for a series circuit is equal to the first resistance plus the second resistance plus the third resistance and so on and so forth for however many resistors there are in series. Plugging in our values, the total resistance for the circuit is equal to ๐‘… one, five ohms, plus ๐‘…, two ohms, plus ๐‘… four, four ohms. When we add those three values together, we get an equivalent resistance of 11 ohms. This is the same as having a circuit with only one resistor of value 11 ohms attached to the battery.

Next, weโ€™ll take a look at a combination circuit that can be simplified down to a simple parallel circuit.

Looking at our circuit, we can see that it would be a simple parallel circuit if not for this branch where we have two resistors in series. This means that if we replace those two resistors with one resistor of equivalent resistance, we would have a simple parallel circuit.

Letโ€™s draw out the simplified circuit. In our simplified circuit, we still have resistor one of two ohms and resistor four of four ohms just as we did up above. But this time we have one resistor on the branch where we had resistor ๐‘… two and ๐‘… three in series. This means that our one resistor has an equivalent resistance of the two resistors in series.

To determine the equivalent resistance of those two resistors in series, we need to use the equation to find total resistance in a series circuit. On the left side of the equation, we keep ๐‘… equivalent for the total resistance of our branch. On the right side equation, we put one ohm in for the value of ๐‘… two and three ohms in for the value of ๐‘… three. When we add one ohm plus three ohm, we get four ohms. This means that resistor ๐‘… has a value of four ohms, which is the equivalent resistance of ๐‘… two and ๐‘… three in series. We use the variable ๐‘… so that we donโ€™t get confused when we go to solve for the equivalent resistance of the entire circuit.

Now, we have a simple parallel circuit of three resistors: ๐‘… one, ๐‘…, and ๐‘… four. We can use the equation for a parallel circuit to determine the total resistance. The left side of the equation stays the same with one over ๐‘… equivalent. On the right side of the equation, we have one over two ohms for our first resistor, one over four ohms for our second resistor, and one over four ohms for our third resistor.

When weโ€™re adding fractions, we need to use the least common denominator. In this case, it would be four ohms. This means that we need to multiply one over two ohms by two over two. We can now add two over four ohms plus one over four ohms plus one over four ohms, which gives us a sum of four over four ohms and can be simplified down to one over one ohm. We multiply both sides of the equation by one ohm and ๐‘… equivalent so that ๐‘… equivalent cancels out on the left side of the equation and one ohm cancels out on the right side of the equation. Leaving us with an equivalent resistance for the circuit of one ohm. This is the same thing as having a circuit with only one resistor of value one ohm attached to the battery.

To determine any other characteristics of a combination circuit, we need to use Ohmโ€™s law. Ohmโ€™s law is an equation that relates the potential difference across the resistor ๐‘‰ to the current through the resistor ๐ผ and the value of the resistance ๐‘…. If we wanna find the total current through each of our combination circuits, we need to use the potential difference of each of the batteries as well as the equivalent resistance of each of the circuits.

To find the current using Ohmโ€™s law, we need to divide both sides of the equation by ๐‘…. On the right side of the equation, the ๐‘…s would cancel out. And the left side of the equation, weโ€™d be left with ๐‘‰ divided by ๐‘…. We can apply this new variation of the equation to both circuits to determine the total current in each.

For our top circuit, the potential of the battery is 12 volts and the equivalent resistance of the circuit is one ohm. 12 volts divided by one ohms is 12 amps. This means that the total current in our circuit is 12 amps. In our bottom circuit, the potential difference of the battery is 22 volts and the equivalent resistance of the circuit is 11 ohms. 22 volts divided by 11 ohms is two amps. The total current through this circuit is two amps.

The current that we found in both of these circuits is the total current not the current through each resistor. We can also use Ohmโ€™s law to determine the potential difference across one of our resistors. If we wanna determine the potential difference across the four-ohm resistor, we need to know the current going through it as well as the value of its resistance.

The current going through the four-ohm resistor would be two amps. We know this because the four-ohm resistor is in series with the other components of the circuit. In a series circuit, the current doesnโ€™t split. Therefore, the total current goes through each of the individual components that are in series. We found that the total current was two amps. Therefore, the current through the four-ohm resistor is also two amps. The resistance value is four ohms. When we multiply two amps by four ohms, we get a potential difference across our four-ohm resistor of eight volts.

Another approach to solve any of these problems when equivalent resistance wonโ€™t work is to apply Kirchoffโ€™s two laws. Letโ€™s refresh our memory on these two laws before we apply them to our previous circuits.

Kirchoffโ€™s first law states that the current into a junction equals the current out of a junction. Applying this law to a parallel circuit when the current splits across multiple paths so that the current in one path is ๐ผ one, the current in the second path is ๐ผ two, the current in the third path is ๐ผ three, and so on. Then adding all those currents together gives us the total current ๐ผ ๐‘‡. Applying this law to a series circuit where the current doesnโ€™t split, the total current will be equal to the current through each of the resistors: ๐ผ one for resistor one, ๐ผ two for resistor two, ๐ผ three for resistor three, and so on.

Kirchoffโ€™s second law states that the sum of all voltages around any closed loop in a circuit must equal zero. We can apply this law to a series circuit because the potential difference is shared or split across components in series. This tells us that for a single battery circuit, the potential difference across all the resistors added together โ€” ๐‘‰ one for resistor one, ๐‘‰ two for resistor two, ๐‘‰ three for resistor three, and so on โ€” is equal to the potential difference of the battery, ๐‘‰ ๐‘‡.

In a simple parallel circuit, this is why the resistor on each branch has the same potential difference as the battery. If we wanna know the current through the three-ohm resistor, we need to apply Kirchoffโ€™s second law to determine the potential difference across the branch. If we follow the path that comes from the battery to the five-ohm resistor up to the three-ohm resistor down to the four-ohm resistor and back to the battery, we can then apply Kirchoffโ€™s second law to this path.

The total potential difference is the potential difference of the battery, 22 volts. ๐‘‰ one is the potential difference across the five-ohm resistor. ๐‘‰ two is the potential difference across the three-ohm resistor. And ๐‘‰ three is the potential difference across the four-ohm resistor. Since we do not know the potential difference across any of the resistors, we will need to plug in for each one Ohmโ€™s law.

For each potential difference, we plug in ๐ผ๐‘…. For the five-ohm resistor, we have an ๐ผ of two amps and an ๐‘… of five ohms. For the three-ohm resistor, we donโ€™t know the current, so we leave it as ๐ผ. And the resistance is three ohms. For the four-ohm resistor, we plug in two amps for the current and four ohms for the resistance.

Simplifying through distribution, two times five is 10, and two times four is eight. To isolate the current, we need to subtract eight volts from both sides and 10 volts from both sides. This will cancel out the eight volts and the 10 volts on the right side of the equation. Subtracting 18 volts from 22 volts leaves us with the left side of the equation of four volts. Our last step is divide both sides of the equation by three ohms, canceling out the three ohms on the right side of the equation. This leaves us with a current of four-thirds amps.

If we apply Kirchoffโ€™s first law, we could find the current through the six-ohm resistor. In this case, the total current that comes into the junction, ๐ผ ๐‘‡, must be equal to the current that goes through the three-ohm resistor plus the current that split to go through the six-ohm resistor. To determine our unknown current, we need to subtract the current through the three-ohm resistor from both sides of the equation. This will cancel out the current for the three-ohm resistor on the right side of the equation.

When we subtract the current through the three-ohm resistor from the total current, we will get the current that goes through the six-ohm resistor. We plug in two amps for the total current and four-thirds amps for the current through the three-ohm resistor as we just found. When subtracting fractions, we need to make sure that we have the least common denominator. In this case, that would be three. So, we multiply two by three over three, which is the same thing as six-thirds. When we subtract four-thirds amps from six-thirds amps, we get two-thirds amps.

Letโ€™s apply what we just learned about combination circuits to an example.

In the circuit shown, the current takes multiple paths from the positive battery terminal to the negative battery terminal. Find the total resistance of the circuit.

That given diagram shows a combination circuit as resistors are both in parallel as well as in series. Before we simplify our circuit to solve for the total resistance, letโ€™s remind ourselves how to find total resistance in both series and parallel circuits. In a series circuit, the total resistance or equivalent resistance is equal to the sum of each individual resistor, ๐‘… one, ๐‘… two, ๐‘… three, and so on, until all the resistors are accounted for. In a parallel circuit, one over the total resistance or equivalent resistance is equal to one over ๐‘… one plus one over ๐‘… two plus one over ๐‘… three and so on and so forth until all the resistors are accounted for.

Looking at our diagram, we can see that if we could replace our two resistors in parallel with an equivalent resistor, then we would have a circuit that is a simple series circuit. We have drawn a simplified version of the circuit below. We need to determine what value of ๐‘… will be equivalent to the 12-ohm and 18-ohm resistors in parallel. To do this, we need to use the equation to find the total resistance of a parallel circuit.

We used ๐‘… for equivalent resistance, so the left side of the equation becomes one over ๐‘…. The right side of the equation are our two resistors, one over 12 ohms plus one over 18 ohms. When adding fractions, we need to find the least common denominator. The least common denominator for 12 ohms and 18 ohms would be 36 ohms. In order for each of our fractions to have a denominator of 36 ohms, our one-over-12-ohm fraction needs to be multiplied by three over three and our one-over-18-ohm fraction has to be multiplied by two over two. This makes our fractions three over 36 ohms plus two over 36 ohms. When we add three over 36 ohms plus two over 36 ohms, we get five over 36 ohms.

To isolate ๐‘…, we multiply both sides of the equation by 36 ohms and ๐‘…. This cancels out ๐‘… on the left side of the equation and 36 ohms on the right side of the equation, giving us the equation 36 ohms is equal to five ๐‘…. Then, we divide both sides by five, canceling out the five on the right side of the equation, giving us an equivalent resistance of 36 divided by five ohms. In decimal form, this would be 7.2 ohms.

Looking at our simplified circuit, we can see that our three resistors are in series with each other. Therefore, we need to use the series equation to find total resistance. The total resistance of our circuit, ๐‘… equivalent, is equal to 14 ohms, the value of the first resistor, plus 10 ohms, the value of the second resistor, plus 7.2 ohms, the value of the equivalent resistance of the 12- and 18-ohm resistors that were in parallel. When we add our three resistors together, we get a value of 31.2 ohms. All the values in our problem have been given to two significant figures. Therefore, we need to round our total resistance to two significant figures, which gives us a total resistance for our circuit of 31 ohms.

Key Points

Characteristics of circuits containing resistors with both series and parallel combinations can be determined by calculating the equivalent resistance of circuit branches. Kirchoffโ€™s laws can be used to calculate currents in circuit branches where currents cannot be calculated using equivalent resistance methods.

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